Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{-5k + 45}{k^2 + k} \div \dfrac{-2k^2 + 162}{k^3 + 5k^2 + 4k} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-5k + 45}{k^2 + k} \times \dfrac{k^3 + 5k^2 + 4k}{-2k^2 + 162} $ First factor out any common factors. $q = \dfrac{-5(k - 9)}{k(k + 1)} \times \dfrac{k(k^2 + 5k + 4)}{-2(k^2 - 81)} $ Then factor the quadratic expressions. $q = \dfrac {-5(k - 9)} {k(k + 1)} \times \dfrac {k(k + 1)(k + 4)} {-2(k - 9)(k + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-5(k - 9) \times k(k + 1)(k + 4) } {k(k + 1) \times -2(k - 9)(k + 9) } $ $q = \dfrac {-5k(k + 1)(k + 4)(k - 9)} {-2k(k - 9)(k + 9)(k + 1)} $ Notice that $(k - 9)$ and $(k + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-5k(k + 1)(k + 4)\cancel{(k - 9)}} {-2k\cancel{(k - 9)}(k + 9)(k + 1)} $ We are dividing by $k - 9$ , so $k - 9 \neq 0$ Therefore, $k \neq 9$ $q = \dfrac {-5k\cancel{(k + 1)}(k + 4)\cancel{(k - 9)}} {-2k\cancel{(k - 9)}(k + 9)\cancel{(k + 1)}} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $q = \dfrac {-5k(k + 4)} {-2k(k + 9)} $ $ q = \dfrac{5(k + 4)}{2(k + 9)}; k \neq 9; k \neq -1 $